Problem 52 - Elementary and Intermediate Algebra

Question

Evaluate each expression without using a calculator. $$ \ln e^{-10} $$

Step-by-Step Solution

Answer

The value of \( \ln(e^{-10}) \) is \( -10 \).

Step 1: Understand the Natural Logarithm Property

Recall the property of the natural logarithm: If you have \( \ln(a^b) \), it simplifies to \( b \cdot \ln(a) \). This property is essential for simplifying expressions like \( \ln(e^{-10}) \).

Step 2: Apply the Property to Simplify the Expression

Using the property \( \ln(a^b) = b \cdot \ln(a) \), rewrite \( \ln(e^{-10}) \) as \( -10 \cdot \ln(e) \).

Step 3: Evaluate \( \ln(e) \)

We know that \( \ln(e) = 1 \) because the natural logarithm of \( e \), the base of natural logarithms, is 1. Thus, we can simplify \( -10 \cdot \ln(e) \) to \( -10 \cdot 1 \).

Step 4: Calculate the Final Result

Multiply the expression to get \( -10 \cdot 1 = -10 \). This is the simplified value of \( \ln(e^{-10}) \).

Key Concepts

Logarithmic Properties

Logarithms have unique properties that help us simplify complex expressions. These properties are invaluable in mathematics, especially when dealing with exponential functions.

• **Product Property**: For two numbers, the logarithm of their product is the sum of the logarithms of those numbers. Mathematically, it is expressed as \( \log_b(xy) = \log_b(x) + \log_b(y) \).
• **Quotient Property**: The logarithm of a division of two numbers equals the difference of logarithms; this is \( \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) \).
• **Power Property**: Crucial for our exercise, it states that if you have something like \( \ln(a^b) \), you can simplify it to \( b \cdot \ln(a) \). Thus, \( \ln(e^{-10}) \) becomes \( -10 \cdot \ln(e) \).

Understanding these properties, especially the Power Property, is essential in logarithmic algebra. It allows for more manageable computations and better insights into the behavior of logarithmic and exponential functions.

Exponentials

Exponentials are mathematical functions in which a number is raised to a power, often featuring a constant base raised to a variable exponent. The most common base is Euler's number, \( e \), approximately equal to 2.71828. This base is central to the study of natural logarithms. Exponentials reflect how quantities grow or decay, such as population growth or radioactive decay, where the rate of change is proportional to the current value. The key exponential properties include:

• **Multiplication of Powers**: \( a^m \times a^n = a^{m+n} \), highlighting how powers add when multiplied.
• **Power of a Power**: Takes the form \( (a^m)^n = a^{mn} \), where exponents multiply when raised to another power.
• **Inverses**: For any exponential function \( e^x \), the inverse is the natural logarithm \( \ln(x) \). This relationship is intrinsically linked and simplifies expressions involving both exponentials and logarithms.

These properties simplify evaluating expressions with exponential components, like converting \( \ln(e^{-10})\) to a simple arithmetic operation thanks to their inverse nature.

Evaluation of Expressions

The process of evaluating expressions, particularly those involving natural logarithms and exponentials, generally requires simplifying the expression through known mathematical properties. In step-by-step fashion, we can rewrite expressions in forms that are easier to evaluate:

• Start by identifying the components of the expression and the properties you can use. For example, with \( \ln(e^{-10}) \), you apply the Power Property of logarithms.
• Rewrite the expression using these properties. Substituting \( \ln(e^{-10}) \) with \( -10 \cdot \ln(e) \) illustrates this point.
• Identify known values, such as \( \ln(e) = 1 \), which are fundamental to simplifying further.
• Perform the arithmetic operations to reach the final answer, as we see with the calculation of \(-10 \times 1\), concluding the evaluation.

Each of these steps contributes to a deeper understanding of how to manipulate and evaluate expressions involving logarithms and exponentials. This not only helps in solving homework exercises but also strengthens overall math skills.